A space craft of mass $M$ is moving with velocity $V$ and suddenly explodes into two pieces. A part of it of mass m becomes at rest, then the velocity of other part will be
A space craft of mass $M$ is moving with velocity $V$ and suddenly explodes into two pieces. A part of it of mass m becomes at rest, then the velocity of other part will be
- A
$\frac{{MV}}{{M - m}}$
- B
$\frac{{MV}}{{M + m}}$
- C
$\frac{{mV}}{{M - m}}$
- D
$\frac{{(M + m)V}}{m}$
Similar Questions
Write the equation of total mechanical energy of a body having mass $m$ and stationary at height $H$.
Write the equation of total mechanical energy of a body having mass $m$ and stationary at height $H$.
A light spring of length $20\, cm$ and force constant $2\, kg/cm$ is placed vertically on a table. A small block of mass $1\, kg$. falls on it. The length $h$ from the surface of the table at which the ball will have the maximum velocity is ............... $\mathrm{cm}$
A light spring of length $20\, cm$ and force constant $2\, kg/cm$ is placed vertically on a table. A small block of mass $1\, kg$. falls on it. The length $h$ from the surface of the table at which the ball will have the maximum velocity is ............... $\mathrm{cm}$
Given in Figures are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Given in Figures are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x=0$, in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $\mathrm{x}$ and the velocity is $\mathrm{v}$. At that instant, which of the following options is/are correct?
(image)
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is : $-\frac{m R}{M+m}$.
[$B$] The position of the point mass is : $x=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$.
[$C$] The velocity of the point mass $m$ is : $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
[$D$] The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.
A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x=0$, in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $\mathrm{x}$ and the velocity is $\mathrm{v}$. At that instant, which of the following options is/are correct?
(image)
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is : $-\frac{m R}{M+m}$.
[$B$] The position of the point mass is : $x=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$.
[$C$] The velocity of the point mass $m$ is : $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
[$D$] The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.
- [IIT 2017]
The potential energy function for a particle executing linear simple harmonic motion is given by $V(x)=$ $k x^{2} / 2,$ where $k$ is the force constant of the oscillator. For $k=0.5\; N m ^{-1}$ the graph of $V(x)$ versus $x$ is shown in Figure. Show that a particle of total energy $1 \;J$ moving under this potential must 'turn back" when it reaches $x=\pm 2 m$
The potential energy function for a particle executing linear simple harmonic motion is given by $V(x)=$ $k x^{2} / 2,$ where $k$ is the force constant of the oscillator. For $k=0.5\; N m ^{-1}$ the graph of $V(x)$ versus $x$ is shown in Figure. Show that a particle of total energy $1 \;J$ moving under this potential must 'turn back" when it reaches $x=\pm 2 m$